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(B) The thermal energy $H$ developed by a resistor of resistance $R$ in time $t$ when a current $I$ flows through it is given by Joule's law of heatin

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$16$

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(B) The thermal energy $H$ developed by a resistor of resistance $R$ in time $t$ when a current $I$ flows through it is given by Joule's law of heating: $H = I^2Rt$.Given,in the first case: $I_1 = 4\,A$,$t = 10\,s$,and energy is $H_1 = H$.So,$H = (4)^2 	imes R 	imes 10 = 160R$.In the second case: $I_2 = 16\,A$,$t = 10\,s$,and let the energy be $H_2$.So,$H_2 = (16)^2 	imes R 	imes 10 = 256 	imes R 	imes 10 = 2560R$.Now,taking the ratio of the two energies:$\frac{H_2}{H} = \frac{2560R}{160R} = 16$.Therefore,$H_2 = 16H$.

The $H$ amount of thermal energy is developed by a resistor in $10\,s$ when a current of $4\,A$ is passed through it. If the current is increased to $16\,A$,the thermal energy developed by the resistor in $10\,s$ will be $........\,H$

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